[SOLVED] Integral formulas

**arbolis** · July 6th 2008, 09:24 AM

Hi MHF!
I need your help for some questions.
First, what is the method to find out that $\int \frac{dx}{1+x^2}=arctan(x)+C$ ? I find it stupid to learn it by heart. There are plenty of other formulas to know, so I would like to know how to recover them without learning all of them.
Second, I have to find $\int \frac{dx}{\sqrt{x^2-9}}$ in term of trigonometric or hyperbolic functions. Looking at my book, I found out that it is equal to $ln|x+\sqrt{x^2-9}|+C$ . Which recalls me of an inverse of $cosh^{-1}$ but cannot find exactly to what it's equal, so I'd appreciate help for this one.
And finally $\int \frac{dx}{\sqrt{4-x^2}}=arcsin\big(\frac{x}{2}\big)+C$ according to my book. But once again, how can I know it's worth an inverse of the sine function? I hope I don't have to learn this by heart.

**galactus** · July 6th 2008, 09:38 AM

Here is a way to look at it and probably the method commonly used.

$\frac{d}{dx}[tan^{-1}(x)]=\frac{1}{1+x^{2}}$

$y=tan^{-1}(x), \;\ x=tan(y)$

Differentiating the right one implicitly gives us:

$\frac{d}{dx}[x]=\frac{d}{dx}[tan(y)]$

$1=sec^{2}(y)\cdot\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2 }(tan^{1}(x))}.....[1]$

Using the triangle we see that:

$sec^{2}(tan^{-1}(x))=1+x^{2}$

And [1] simplifies to $\frac{dy}{dx}=\frac{1}{1+x^{2}}$

Therefore, $\int\frac{1}{1+x^{2}}=tan^{-1}(x)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

To find $\int\frac{1}{\sqrt{x^{2}-9}}dx$

let $x=3sec{\theta}, \;\ dx=3sec{\theta}tan{\theta}d{\theta}$

When you make the subs and simplify it whittles down to

$\int sec{\theta}d{\theta}$

Then, to get back in terms of x just resub ${\theta}=sec^{-1}(\frac{x}{3})$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~

$\int\frac{1}{\sqrt{4-x^{2}}}dx$

This one simplifies wonderfully. Let $x=2sin{\theta}, \;\ dx=2cos{\theta}d{\theta}$

Make the subs and you finally get:

$\int d{\theta}={\theta}$

What is theta?. ${\theta}=sin^{-1}(x/2)$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~

To tell when to use what, just remember the formulas:

$\sqrt{a^{2}-x^{2}}$ , then use $x=a\cdot sin{\theta}$

$\sqrt{a^{2}+x^{2}}$ , then use $x=a\cdot tan{\theta}$

$\sqrt{x^{2}-a^{2}}$ , then use $x=a\cdot sec{\theta}$

Can you see whey we use those when we do?. Because they result in the proper identities that get rid of the radical.

**arbolis** · July 6th 2008, 09:53 AM

Thanks a lot, I'll be analyzing this closely.
You made an error of typo in the line " $\frac{dy}{dx}=\frac{1}{sec^{2}(y)}=\frac{1}{sec^{2 }(tan^{1}(x))}$ .....[1]" Must be $tan^{-1}$ . Otherwise I think all is O.K. Thanks so much!!!

**Reckoner** · July 6th 2008, 10:01 AM

Originally Posted by arbolis

And finally $\int \frac{dx}{\sqrt{4-x^2}}=arcsin\big(\frac{x}{2}\big)+C$ according to my book. But once again, how can I know it's worth an inverse of the sine function? I hope I don't have to learn this by heart.

Let $y = \arcsin x$ . Then we have

$x = \sin y$

$\Rightarrow \cos y\frac{dy}{dx} = 1$

$\Rightarrow \frac{dy}{dx} = \frac1{\cos y} = \frac1{\sqrt{1 - \sin^2 y}} = \frac1{\sqrt{1 - x^2}}$

The other inverse functions are similar.

**nikhil** · July 6th 2008, 11:22 AM

hi!arbolis,hope ur having a great time.
Hmmmmm..... You have asked a very basic question.though i can tell you how to integrate some function by a definite procedure (like for your first integral question put x=tany,for second put X=3secy etc) but i think i should give you basics that are behind the procedures
The main problem is that unlike differentiation there is no definite first principle for integration.when we perform integration,we take it as a reverse process of differentiation. Like
d(sinx)/dx=cosx
therefor
integral cosxdx=sinx.
So as you can see if you know the derivative of a function then the integral of its derivative is the function itself.so whenever we need to integrate a complex looking function we try to make it less complex by substitution.
Like A= $\int \frac{dx}{1+x^2}$
let x=tany
differentiating both sides we get
dx=(secy)^2
now substituting these value we get
A= $<br /> \int\frac{(secy)^{2} dy}{1+(tany)^2}<br />$
but 1+(tany)^2=(secy)^2
so A= $<br /> \int dy<br />$ [so because of the substitution integral became easy]
so A=y+c
since x=tany
y=arctanx
so A=arctanx+c
i hope this will help in widening your thinking.
REMEMBER it may not be possible to solve all questions by substitution.so like differentiation there are some other procedures[like partial fraction method}.

**ThePerfectHacker** · July 6th 2008, 12:25 PM

It is really helpful to memorize some basic forms. For example the derivative of $\sin^{-1}$ (on $(-1,1)$ ). The way we do this is to noting that $\sin (\sin^{-1} x ) = x$ thus $(\sin^{-1} x) \cos (\sin^{-1} x) = 1$ by the chain rule. But, $\cos (\sin^{-1} x) = \sqrt{1-x^2}$ which means $(\sin^{-1} x) ' = \tfrac{1}{\sqrt{1-x^2}}$ . Here is an example, $\smallint \tfrac{dx}{\sqrt{9-x^2}}$ . We can factor this, $\tfrac{1}{3}\smallint \tfrac{dx}{\sqrt{1-(\frac{x}{3})^2}}$ . Now let $t=\tfrac{x}{3}$ and we get $\smallint \tfrac{dt}{\sqrt{1-t^2}} = \sin^{-1} t + k = \sin^{-1} \tfrac{x}{3}+k$ .

Now find a similar form for $\sinh^{-1} x$ . Again, $\sinh (\sinh^{-1} x) = x$ by the chain rule we get $(\sinh^{-1} x)' \cosh(\sinh^{-1} x) = 1$ . Now note that $\cosh(\sinh^{-1} x) = \sqrt{x^2+1}$ *. For example, $\smallint \tfrac{dx}{\sqrt{x^2+4}}$ can be solved in a similar way as above problem. Inside of doing all those substitutions.

*)Let $f(x) = \cosh(\sinh^{-1} x)$ and see that $f(x)>0$ . But then $f^2(x) = \cosh^2 (\sinh^{-1} x) = 1 + \sinh^2(\sinh^{-1} x) = 1+x^2$ . Thus, $f(x) = \sqrt{x^2+1}$ .

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